How to find the sum of an array of numbers

Question

Given an array [1, 2, 3, 4], how can I find the sum of its elements? (In this case, the sum would be 10.)

I thought $.each might be useful, but I'm not sure how to implement it.

Answer 1

This'd be exactly the job for reduce.

If you're using ECMAScript 2015 (aka ECMAScript 6):

const sum = [1, 2, 3].reduce((partialSum, a) => partialSum + a, 0);
console.log(sum); // 6

DEMO

For older JS:

const sum = [1, 2, 3].reduce(add, 0); // with initial value to avoid when the array is empty

function add(accumulator, a) {
  return accumulator + a;
}

console.log(sum); // 6

Isn't that pretty? :-)

Answer 2

Recommended (reduce with default value)

Array.prototype.reduce can be used to iterate through the array, adding the current element value to the sum of the previous element values.

console.log(
  [1, 2, 3, 4].reduce((a, b) => a + b, 0)
)
console.log(
  [].reduce((a, b) => a + b, 0)
)

Without default value

You get a TypeError

console.log(
  [].reduce((a, b) => a + b)
)

Prior to ES6's arrow functions

console.log(
  [1,2,3].reduce(function(acc, val) { return acc + val; }, 0)
)

console.log(
  [].reduce(function(acc, val) { return acc + val; }, 0)
)

Non-number inputs

If non-numbers are possible inputs, you may want to handle that?

console.log(
  ["hi", 1, 2, "frog"].reduce((a, b) => a + b)
)

let numOr0 = n => isNaN(n) ? 0 : n

console.log(
  ["hi", 1, 2, "frog"].reduce((a, b) => 
    numOr0(a) + numOr0(b))
)

Non-recommended dangerous eval use

We can use eval to execute a string representation of JavaScript code. Using the Array.prototype.join function to convert the array to a string, we change [1,2,3] into "1+2+3", which evaluates to 6.

console.log(
  eval([1,2,3].join('+'))
)

//This way is dangerous if the array is built
// from user input as it may be exploited eg: 

eval([1,"2;alert('Malicious code!')"].join('+'))

Of course displaying an alert isn't the worst thing that could happen. The only reason I have included this is as an answer Ortund's question as I do not think it was clarified.

Answer 3

Why not reduce? It's usually a bit counter intuitive, but using it to find a sum is pretty straightforward:

var a = [1,2,3];
var sum = a.reduce(function(a, b) { return a + b; }, 0);

Answer 4

var arr = [1, 2, 3, 4];
var total = 0;
for (var i in arr) {
  total += arr[i];
}

Answer 5

var total = 0;
$.each(arr,function() {
    total += this;
});

Answer 6

Anyone looking for a functional oneliner like me?

Assuming:

const arr = [1, 2, 3, 4];

Here's the oneliner for modern JS:

sum = arr.reduce((a, b) => a + b, 0);

(If you happen to have to support ye olde IE without arrow functions:)

sum = arr.reduce(function (a, b) {return a + b;}, 0);

Note that 0 is the initial value here, so you can use that as offset if needed. Also note that this initial value is needed, otherwise calling the function with an empty array will error.

Answer 7

If you happen to be using Lodash you can use the sum function

array = [1, 2, 3, 4];
sum = _.sum(array); // sum == 10

Answer 8

This is possible by looping over all items, and adding them on each iteration to a sum-variable.

var array = [1, 2, 3];

for (var i = 0, sum = 0; i < array.length; sum += array[i++]);

JavaScript doesn't know block scoping, so sum will be accesible:

console.log(sum); // => 6

The same as above, however annotated and prepared as a simple function:

function sumArray(array) {
  for (
    var
      index = 0,              // The iterator
      length = array.length,  // Cache the array length
      sum = 0;                // The total amount
      index < length;         // The "for"-loop condition
      sum += array[index++]   // Add number on each iteration
  );
  return sum;
}

Answer 9

arr.reduce(function (a, b) {
    return a + b;
});

Reference: Array.prototype.reduce()

Answer 10

OK, imagine you have this array below:

const arr = [1, 2, 3, 4];

Let's start looking into many different ways to do it as I couldn't find any comprehensive answer here:

1) Using built-in reduce()

function total(arr) {
  if(!Array.isArray(arr)) return;
  return arr.reduce((a, v)=>a + v);
}

2) Using for loop

function total(arr) {
  if(!Array.isArray(arr)) return;
  let totalNumber = 0;
  for (let i=0,l=arr.length; i<l; i++) {
     totalNumber+=arr[i];
  }
  return totalNumber;
}

3) Using while loop

function total(arr) {
  if(!Array.isArray(arr)) return;
  let totalNumber = 0, i=-1;
  while (++i < arr.length) {
     totalNumber+=arr[i];
  }
  return totalNumber;
}

4) Using array forEach

function total(arr) {
  if(!Array.isArray(arr)) return;
  let sum=0;
  arr.forEach(each => {
    sum+=each;
  });
  return sum;
};

and call it like this:

total(arr); //return 10

It's not recommended to prototype something like this to Array...

Answer 11

Funny approach:

eval([1,2,3].join("+"))

Answer 12

You can also use reduceRight.

[1,2,3,4,5,6].reduceRight(function(a,b){return a+b;})

which results output as 21.

Reference: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/ReduceRight

Answer 13

A standard JavaScript solution:

var addition = [];
addition.push(2);
addition.push(3);

var total = 0;
for (var i = 0; i < addition.length; i++)
{
    total += addition[i];
}
alert(total);          // Just to output an example
/* console.log(total); // Just to output an example with Firebug */

This works for me (the result should be 5). I hope there is no hidden disadvantage in this kind of solution.

Answer 14

I am a beginner with JavaScript and coding in general, but I found that a simple and easy way to sum the numbers in an array is like this:

    var myNumbers = [1,2,3,4,5]
    var total = 0;
    for(var i = 0; i < myNumbers.length; i++){
        total += myNumbers[i];
    }

Basically, I wanted to contribute this because I didn't see many solutions that don't use built-in functions, and this method is easy to write and understand.

Answer 15

Use a for loop:

const array = [1, 2, 3, 4];
let result = 0;

for (let i = 0; i < array.length - 1; i++) {
  result += array[i];
}

console.log(result); // Should give 10

Or even a forEach loop:

const array = [1, 2, 3, 4];
let result = 0;

array.forEach(number => {
  result += number;
})

console.log(result); // Should give 10

For simplicity, use reduce:

const array = [10, 20, 30, 40];
const add = (a, b) => a + b
const result = array.reduce(add);

console.log(result); // Should give 100

Answer 16

var totally = eval(arr.join('+'))

That way you can put all kinds of exotic things in the array.

var arr = ['(1/3)','Date.now()','foo','bar()',1,2,3,4]

I'm only half joking.

Answer 17

ES6 for..of

let total = 0;

for (let value of [1, 2, 3, 4]) {
    total += value; 
}

Answer 18

A short piece of JavaScript code would do this job:

var numbers = [1,2,3,4];
var totalAmount = 0;

for (var x = 0; x < numbers.length; x++) {

    totalAmount += numbers[x];
}

console.log(totalAmount); //10 (1+2+3+4)

Answer 19

Use reduce

let arr = [1, 2, 3, 4];

let sum = arr.reduce((v, i) => (v + i));

console.log(sum);

Answer 20

A few people have suggested adding a .sum() method to the Array.prototype. This is generally considered bad practice so I'm not suggesting that you do it.

If you still insist on doing it then this is a succinct way of writing it:

Array.prototype.sum = function() {return [].reduce.call(this, (a,i) => a+i, 0);}

then: [1,2].sum(); // 3

Note that the function added to the prototype is using a mixture of ES5 and ES6 function and arrow syntax. The function is declared to allow the method to get the this context from the Array that you're operating on. I used the => for brevity inside the reduce call.

Answer 21

No need to initial value! Because if no initial value is passed, the callback function is not invoked on the first element of the list, and the first element is instead passed as the initial value. Very cOOl feature :)

[1, 2, 3, 4].reduce((a, x) => a + x) // 10
[1, 2, 3, 4].reduce((a, x) => a * x) // 24
[1, 2, 3, 4].reduce((a, x) => Math.max(a, x)) // 4
[1, 2, 3, 4].reduce((a, x) => Math.min(a, x)) // 1

Answer 22

Here's an elegant one-liner solution that uses stack algorithm, though one may take some time to understand the beauty of this implementation.

const getSum = arr => (arr.length === 1) ? arr[0] : arr.pop() + getSum(arr);

getSum([1, 2, 3, 4, 5]) //15

Basically, the function accepts an array and checks whether the array contains exactly one item. If false, it pop the last item out of the stack and return the updated array.

The beauty of this snippet is that the function includes arr[0] checking to prevent infinite looping. Once it reaches the last item, it returns the entire sum.

Answer 23

Accuracy

Sort array and start sum form smallest numbers (snippet shows difference with nonsort)

[...arr].sort((a,b)=>a-b).reduce((a,c)=>a+c,0)

arr=[.6,9,.1,.1,.1,.1]

sum     =                       arr.reduce((a,c)=>a+c,0)
sortSum = [...arr].sort((a,b)=>a-b).reduce((a,c)=>a+c,0)

console.log('sum:     ',sum);
console.log('sortSum:',sortSum);
console.log('sum==sortSum :', sum==sortSum);

// we use .sort((a,b)=>a-b) instead .sort() because
// that second one treat elements like strings (so in wrong way)
// e.g [1,10,9,20,93].sort() -->  [1, 10, 20, 9, 93]

For multidimensional array of numbers use arr.flat(Infinity)

arr= [ [ [1,2,3,4],[1,2,3,4],[1,2,3,4] ],
       [ [1,2,3,4],[1,2,3,4],[1,2,3,4] ] ];
      
sum = arr.flat(Infinity).reduce((a,c)=> a+c,0);

console.log(sum);  // 60

Answer 24

You can try the following code:

[1, 2, 3, 4].reduce((pre,curr)=>pre+curr,0)

Answer 25

Those are really great answers, but just in case if the numbers are in sequence like in the question ( 1,2,3,4) you can easily do that by applying the formula (n*(n+1))/2 where n is the last number

Answer 26

You can combine reduce() method with lambda expression:

[1, 2, 3, 4].reduce((accumulator, currentValue) => accumulator + currentValue);

Answer 27

With reduce()

[1, 2, 3, 4].reduce((a, b) => a + b, 0); // 10

With forEach()

let sum = 0;
[1, 2, 3, 4].forEach(n => sum += n);
sum; // 10

With Parameter

function arrSum(arr) { 
  sum = 0;  
  arr.forEach(n => sum += n); 
  return sum; 
}

arrSum([1, 2, 3, 4]) // 10

Answer 28

i saw all answers going for 'reduce' solution

var array = [1,2,3,4]
var total = 0
for (var i = 0; i < array.length; i++) {
    total += array[i]
}
console.log(total)

Answer 29

very simple

step 1 we should have an array like :

const arrayNumber = [500,152,154,1555,12445];

step 2 (you can ignore this step if) step is to be sur that all values in table are number for that

let newArray = [];
for (let i = 0; i < arrayNumber.length; i++) {
        newArray.push(parseInt(arrayNumber[i], 10));
      }

step 3

const sumInArray = dataData.reduce( (a, b) => a + b);

finally

console.log(sumInArray);

Answer 30

Simplest answer to understand underlying process:

let array = [10, 20, 30, 40, 50]
let total = 0

for(let i in array)
{
    total += array[i]
}

console.log(total)

& if you're already familiar with underlying process then built-in method can save you time:

let array = [10, 20, 30, 40, 50]
let total = array.reduce((x, y) => x + y)
console.log(total)

Answer 31

Cool tricks here, I've got a nit pick with a lot of the safe traditional answers not caching the length of the array.

function arraySum(array){
  var total = 0,
      len = array.length;

  for (var i = 0; i < len; i++){
    total += array[i];
  }

  return total;
};

var my_array = [1,2,3,4];

// Returns 10
console.log( arraySum( my_array ) );

Without caching the length of the array the JS compiler needs to go through the array with every iteration of the loop to calculate the length, it's unnecessary overhead in most cases. V8 and a lot of modern browsers optimize this for us, so it is less of a concern then it was, but there are older devices that benefit from this simple caching.

If the length is subject to change, caching's that could cause some unexpected side effects if you're unaware of why you're caching the length, but for a reusable function who's only purpose is to take an array and add the values together it's a great fit.

Here's a CodePen link for this arraySum function. http://codepen.io/brandonbrule/pen/ZGEJyV

It's possible this is an outdated mindset that's stuck with me, but I don't see a disadvantage to using it in this context.

Answer 32

Object.defineProperty(Object.prototype, 'sum', {
    enumerable:false,
    value:function() {
        var t=0;for(var i in this)
            if (!isNaN(this[i]))
                t+=this[i];
        return t;
    }
});

[20,25,27.1].sum()                 // 72.1
[10,"forty-two",23].sum()          // 33
[Math.PI,0,-1,1].sum()             // 3.141592653589793
[Math.PI,Math.E,-1000000000].sum() // -999999994.1401255

o = {a:1,b:31,c:"roffelz",someOtherProperty:21.52}
console.log(o.sum());              // 53.519999999999996

Answer 33

This is much easier

function sumArray(arr) {
    var total = 0;
    arr.forEach(function(element){
        total += element;
    })
    return total;
}

var sum = sumArray([1,2,3,4])

console.log(sum)

Answer 34

With ES6 rest parameter

let array = [1, 2, 3, 4]

function sum(...numbers) {
    let total = 0;
    for (const number of numbers) {
        total += number;
    }
    return total;
}

console.log(sum(...array));

Answer 35

A simple method example:

function add(array){
    var arraylength = array.length;
    var sum = 0;
    for(var timesToMultiply = 0; timesToMultiply<arraylength; timesToMultiply++){
        sum += array[timesToMultiply];
    }

    return sum;
}

console.log(add([1, 2, 3, 4]));

Answer 36

When the array consists of strings one has to alter the code. This can be the case, if the array is a result from a databank request. This code works:

alert(
["1", "2", "3", "4"].reduce((a, b) => Number(a) + Number(b), 0)
);

Here, ["1", "2", "3", "4"] ist the string array and the function Number() converts the strings to numbers.

Answer 37

Is there a reason not to just filter the array first to remove non-numbers? Seems simple enough:

[1, 2, 3, null, 'a'].filter((x) => !isNaN(x)).reduce((a, b) => a + b)

Answer 38

try this...

function arrSum(arr){
    total = 0;  
    arr.forEach(function(key){
        total = total + key;            
    });
    return total;
}

Answer 39

Vanilla JavaScript is all you need:

> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; a.forEach(function(e){sum += e}); sum
10
> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; a.forEach(e => sum += e); sum
10
> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; for(e in a) sum += e; sum
"00123foobar"
> a = [1,2,3,4]; a.foo = 5; a['bar'] = 6; sum = 0; for(e of a) sum += e; sum
10

Answer 40

This function can sum up all the numbers -

 function array(arr){
   var sum = 0;
   for (var i = 0; i< arr.length; i++){
    sum += arr[i];
   }
   console.log(sum);
 }
 array([5, 1, 3, 3])

Answer 41

Considering you have the array

const arr = [1, 2, 3, 4];

For me, the most intuitive way is using a for-in

let sum = 0;
for(var value in arr) {
    sum += arr[value];
}

console.log('Array:', arr);
console.log('Sum:', sum);

Yet, you can also use Arrow Functions and the reduce function

const sum = arr.reduce(function (a, b) {
    return a + b;
}, 0);

console.log('Array:', arr);
console.log('Sum:', sum);

They're both gonna output

Array: [ 1, 2, 3, 4]
Sum: 10

Answer 42

getTotal = (arr) => {
    let total = 0
    for (let i = 0; i < arr.length; i++) {
        total += arr[i];
    }
    return total
}

getTotal([1, 2, 3, 4]) // 10
getTotal([1, 2, 3, 4, 5]) // 15

Answer 43

Just use this function:

function sum(pArray)
{
  pArray = pArray.reduce(function (a, b) {
      return a + b;
  }, 0);

  return pArray;
}

function sum(pArray)
{
  pArray = pArray.reduce(function (a, b) {
      return a + b;
  }, 0);

  return pArray;
}

var arr = [1, 4, 5];

console.log(sum(arr));

Answer 44

A "duplicate" question asked how to do this for a two-dimensional array, so this is a simple adaptation to answer that question. (The difference is only the six characters [2], 0, which finds the third item in each subarray and passes an initial value of zero):

const twoDimensionalArray = [
  [10, 10, 1],
  [10, 10, 2],
  [10, 10, 3],
];
const sum = twoDimensionalArray.reduce( (partial_sum, a) => partial_sum + a[2], 0 ) ; 
console.log(sum); // 6

Answer 45

You can try this:

var arr = [100,114,250,1200];
var total = 0; 
for(var i in arr){
  total += parseInt(arr[i]);
}

console.log(total);

Output will be: 1664

Or if value is Float, then try this:

var arr = [100.00,114.50,250.75,1200.00];
    var total = 0; 
    for(var i in arr){
      total += parseFloat(arr[i]);
    }
    
    console.log(total.toFixed(2));

Output will be: 1665.25

Answer 46

I´m aware that this is an old post. But I just wanted to share my approach to the problem.

let myArr = [1, 3, 4, 5, 6, 7, 8];
let counter = 0;
for(let i = 0; i < myArr.length; i++){
  counter = counter + myArr[i];
    console.log(counter);}

Answer 47

//Try this way

const arr = [10,10,20,60]; 
const sumOfArr = (a) =>{
    let sum=0;
    for(let i in a) { 
        sum += a[i];
    }
    return sum;
}
console.log(sumOfArr(arr))

Answer 48

The truth is there is an old and funny classic solution (beside newbie 'foreach' and 'reduce'): the classic for of.

y = 0;
for (x of [1, 2, 3, 4]) y+=x;

Answer 49

Someone also can use map() function for the summation of an array values.

function sumOfArrVal(arr){
    let sum=0;
    arr.map(val=>sum+=val)
    return sum
}

let result=sumOfArrVal([1,2,3,4])
console.log(result)

Answer 50

No one mentioned functional programming, but it would be very clean approach to use Ramda in this case:

//Assuming you use nodejs, but can also be used in browser
const R = require('ramda');

let nums = [2, 4, 6, 8, 10];
console.log(R.sum(nums));

Answer 51

I see that this is an old question, but hardly any of the answers actually use jQuery and specifically $.each as requested by the asker. There is a way to do this, but the native approaches as suggested by many (i.e. using Array.reduce()) are much better and easier.

Here's the jQuery way if you absolutely have to use it:

  var arr = [1,2,3,4];
  var sum = 0;
  
  $.each( arr, (i, v) => sum += v);

  alert(sum); // This will be 10

Answer 52

Use recursion

var sum = (arr) => arr.length === 1 ? arr[0] : arr.shift() + sum(arr);
sum([1,2,3,4]) // 10

Answer 53

In addition, sum with es6 for simple array.

const sum = [1, 2, 3].reduce((partial_sum, a) => partial_sum + a,0);
 
console.log(sum); 

For object array with default initialize value

const totalAmount = obj => 
    Object.values(obj).reduce((acc, { order_qty, mrp_price }) => 
    acc + order_qty * mrp_price, 0);
    
    console.log(totalAmount); 

Answer 54

For really large numbers cycling or reducing could be process intensive. What about using Gauss?

sum = (n * (n+1))/2;

From mathcentral.

Answer 55

    <!DOCTYPE html>
    <html>
    <body>

      <p>Click the button to join two arrays.</p>
      <button onclick="myFunction()">Try it</button>
      <p id="demo"></p>
    <script>
var hege = [1, 2,4,6,7,8,8];
var stale = [1, 2,4,5];
function myFunction() {
    console.log((hege.length > stale.length))    
    var children  = (hege.length > stale.length)? abc1() :abc2();       document.getElementById("demo").innerHTML = children;
}
function abc1(){
    console.log(hege,"Abc1")    
    var abcd=hege.map(function (num, idx) {
        console.log(hege.length , idx)
        return stale.length>idx?num + stale[idx]:num;
    })
    return abcd;
}

function abc2(){

    console.log(hege,"Abc2",stale)    
    var abcd=stale.map(function (num, idx) {
        console.log(hege.length , idx)
        return hege.length>idx?num + hege[idx]:num;
    })
    return abcd;
}
</script>

</body>
</html>

Answer 56

Use map:

var sum = 0;
arr.map(function(item){
    sum += item;
});

// sum now contains the total.

You could potentially add the method to the Array prototype.

Array.prototype.sum = function(){
    var sum = 0;
    this.map(function(item){
        sum += item;
    });
    return sum;
}

Then you can use it on any Array like so:

arr.sum();